Question

You digest a random piece of linear DNA of 2222 bp long with MluCl (5’ AATT3’).
a. Approximately how many cuts and how fragments (whole number) would we expect to get? (2 pt)
b. If the random piece of DNA from a) was subsequently found to have a G/C ratio of 65% would you expect the number of fragments to increase or decrease and why? (1 pt)
c. If the random piece of DNA was actually circular plasmid DNA, how would that affect the number of fragments? (1 pt)

Answer 1

The steps involved here is

1. Calcuation of the probability of restriction sequence and

2. Calculation of the restriction sites in given length of DNA.

a) The answer is

Step 1: Calcuation of the probability of restriction sequence:

In DNA the total number of base pairs are 4 (A, T, G, C), so at once only one base will be incorporeated at a single site out of four, so the probability of each base is 1/4.

So, the probability of site AATT is = 1/4*1/4*1/4*1/4 = 1/256 which means one restriction site may present for every 256 basepairs.

Step 2. Calculation of the restriction sites in 2222 bp length of DNA:

256 > 1

for 2222 > ?

so, the answre is 2222*1/256 = 8.67 ~ 8 sites.

so, one cut in the linear DNA strand makes two fragment, so, 8 sites makes it in to 9 fragments.

b) The answer is

Step 1: Calcuation of the probability of restriction sequence:

In DNA the A+T+G+C = 100%, so, A+T% = 100-G/C % = 100-65 = 35.

G+C = 65% (G = 65/2 = 32.5% and C = 65/2 = 32.5% ).

A+T = 35% (A = 35/2 = 17.5% and T = 35/2 = 17.5% ).

So, the probabilities are

G = 32.5% = 32.5/100= 3.25/10

C = 32.5% = 32.5/100= 3.25/10

A = 17.5% = 17.5/100= 1.75/10

T = 17.5% = 17.5/100= 1.75/10

here, the probabilty of AATT = 1.75/10 * 1.75/10 *1.75/10 *1.75/10 = 9.37/10000

which means ~9 restriction sites may present for every 10000 basepairs.

Step 2. Calculation of the restriction sites in 2222 bp length of DNA:

10000 > 9.37

for 2222 > ?

so, the answre is 2222*9.37/10000 = 2.08 ~ 2 sites.

so, one cut in the linear DNA strand makes two fragment, so, 2 sites makes it in to 3 fragments.

c) If one cut makes the linear fragment in two fragments, but where is one cut in circular DNA may truns it in to first linear means one cut makes one fragment in ciruclar DNA, so, 8 restrictions sites make this circular DNA in to 8 fragments and 2 restriction sites make the circular dna in to two fragment.